- (a) Given
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta \quad n \in \mathbb { N }$$
show that
$$32 \cos ^ { 6 } \theta \equiv \cos 6 \theta + 6 \cos 4 \theta + 15 \cos 2 \theta + 10$$
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f237de57-ed6d-4bea-8bb0-1b4e5b66d7da-22_218_357_653_331}
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\caption{Figure 1}
\end{figure}
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f237de57-ed6d-4bea-8bb0-1b4e5b66d7da-22_307_824_621_897}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 1 shows a solid paperweight with a flat base.
Figure 2 shows the curve with equation
$$y = H \cos ^ { 3 } \left( \frac { x } { 4 } \right) \quad - 4 \leqslant x \leqslant 4$$
where \(H\) is a positive constant and \(x\) is in radians.
The region \(R\), shown shaded in Figure 2, is bounded by the curve, the line with equation \(x = - 4\), the line with equation \(x = 4\) and the \(x\)-axis.
The paperweight is modelled by the solid of revolution formed when \(R\) is rotated \(\mathbf { 1 8 0 } ^ { \circ }\) about the \(x\)-axis.
Given that the maximum height of the paperweight is 2 cm ,
(b) write down the value of \(H\).
(c) Using algebraic integration and the result in part (a), determine, in \(\mathrm { cm } ^ { 3 }\), the volume of the paperweight, according to the model. Give your answer to 2 decimal places.
[0pt]
[Solutions based entirely on calculator technology are not acceptable.]
(d) State a limitation of the model.